If $x^2 +\dfrac{1}{x^2} = 7$, find the values of :

(i) $x -\dfrac{1}{x}$

(ii) $x +\dfrac{1}{x}$

(iii) $3x^2-\dfrac{3}{x^2}$

(i) Given, $x^2 +\dfrac{1}{x^2} = 7$

Subtracting 2 from both sides, we get

$⇒ x^2 + \dfrac{1}{x^2} - 2 = 7 - 2\\[1em] ⇒ x^2 + \dfrac{1}{x^2} - 2 \times x^2 \times \dfrac{1}{x^2} = 5\\[1em] ⇒ \Big(x - \dfrac{1}{x}\Big)^2 = 5\\[1em] ⇒ x - \dfrac{1}{x} = \sqrt5 \text{ or } - \sqrt5$

Hence, $x - \dfrac{1}{x} = ± \sqrt5$.

(ii) Given, $x^2 +\dfrac{1}{x^2} = 7$

Adding 2 on both sides, we get

$⇒ x^2 + \dfrac{1}{x^2} + 2 = 7 + 2\\[1em] ⇒ x^2 + \dfrac{1}{x^2} + 2 \times x^2 \times \dfrac{1}{x^2} = 9\\[1em] ⇒ \Big(x + \dfrac{1}{x}\Big)^2 = 9\\[1em] ⇒ x + \dfrac{1}{x} = \sqrt9 \text{ or } - \sqrt9\\[1em] ⇒ x + \dfrac{1}{x} = 3 \text{ or } - 3$

Hence, $x + \dfrac{1}{x} = ± 3$.

(iii) Given, $3x^2-\dfrac{3}{x^2}$

$⇒ 3\Big(x^2 - \dfrac{1}{x^2}\Big) = 3\Big(x - \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x}\Big)\\[1em] = 3(± \sqrt5) \times (± 3)\\[1em] = ± 9\sqrt5\\[1em]$

Hence, $3x^2-\dfrac{3}{x^2} = ± 9\sqrt5$.